Prof. Brian L. Evans

Dept. of Electrical and Computer Engineering

The University of Texas at Austin

EE345S Real-Time Digital Signal Processing Lab Fall 2006

Lecture 14

Digital Pulse Amplitude Modulation

14 - 2

Introduction

• Modulate M = 2J discrete messages or J bits of information into amplitude of signal

• If amplitude mapping changes at symbol rate of fsym, then bit rate is J fsym

• Conventional mapping of discretemessages to M uniformly space amplitudes

• Pulse amplitude modulated (PAM) signal

)12( idai 2 , ... , 0 ..., , 1

2

MMi

) ( )(*sym

kk Tktats

No pulses overlap in time:requires infinite bandwidth

fsym = 1 / Tsym

14 - 3

Pulse Shaping

• Infinite bandwidth cannot be sent in practice• Limit bandwidth by pulse shaping

filter with impulse response gT(t)

• L samples per symbol durationn is symbol index

m is sample index in a symbol: m = 0, 1, 2,…, L-1.

symTk

k Tktgatssym

)(*

symsymsymT

kksymsym kTT

L

mnTgaT

L

mnTs

sym *

mknLgakLmnLgamLnssymsym T

kkT

kk

)( *

At each time t, k is indexed over number of

overlapping pulses

At indices n & m, k is indexed over number of overlapping pulses

14 - 4

Pulse Shaping Example

2-PAM with Raised Cosine Pulse Shaping

14 - 5

Pulse Shaping Block Diagram

• Upsampling by L denoted as LOutpus input sample followed by L-1 zeros

Upsampling by converts symbol rate to sampling rate

• Pulse shaping (FIR) filter gTsym[m]

Fills in zero values generated by upsampler

Multiplies by zero most of time (L-1 out of every L times)

D/A TransmitFilter

ak gTsym[m] L

symbol rate

sampling rate

sampling rate

analog analog

14 - 6

Digital Interpolation Example

• Upsampling by 4 (denoted by 4)Output input sample followed by 3 zeros

Four times the samples on output as input

Increases sampling rate by factor of 4

• FIR filter performs interpolationLowpass filter with stopband frequency stopband / 4

For fsampling = 176.4 kHz, = / 4 corresponds to 22.05 kHz

Digital 4x Oversampling Filter

16 bits44.1 kHz

28 bits176.4 kHz4 FIR Filter16 bits

176.4 kHz

1 2

Input to Upsampler by 4

n

0

n’

Output of Upsampler by 4

1 2 3 4 5 6 7 80

1 2

Output of FIR Filter

3 4 5 6 7 8

n’

0

14 - 7

Pulse Shaping Filter Bank

• Simplify by avoiding multiplication by zeroSplit the long pulse shaping filter into L short polyphase filters

operating at symbol rategTsym,0[n]

gTsym,1[n]

gTsym,L-1[n]

ak

D/A TransmitFilter

s(Ln)

s(Ln+1)

s(Ln+(L-1))

Filter Bank Implementation

D/A TransmitFilter

ak gTsym[m] L

symbol rate

sampling rate

sampling rate

analog analog

14 - 8

Pulse Shaping Filter Bank Example

• Pulse length 24 samples and L = 4 samples/symbol

• Derivation in Tretter's manual,

• Define mth polyphase filter

• Four six-tap polyphase filters (next slide)

symsymsymT

n

nkksymsym kTT

L

mnTgaT

L

mnTs

3

2

* 1,...,1 ,0 Lm

symsymTmT T

L

mnTgng

symsym][,

kngaTL

mnTs mT

n

nkksymsym sym

,

3

2

*

mknLgamnLs T

n

nkk

)( 3

2

* Six pulses contribute to each output sample

1,...,1 ,0 Lm

14 - 9

Pulse Shaping Filter Bank Example

gTsym,0[n]24 samples

in pulse

x marks samples of polyphase

filter

4 samples per symbol

Polyphase filter 0 response is the first sample of the

pulse shape plus every fourth sample after that

Polyphase filter 0 has only one non-zero sample.

14 - 10

Pulse Shaping Filter Bank Example24 samples

in pulse

x marks samples of polyphase

filter

4 samples per symbol

Polyphase filter 1 response is the second sample of the

pulse shape plus every fourth sample after that

gTsym,1[n]

14 - 11

Pulse Shaping Filter Bank Example24 samples

in pulse

x marks samples of polyphase

filter

4 samples per symbol

Polyphase filter 2 response is the third sample of the

pulse shape plus every fourth sample after that

gTsym,2[n]

14 - 12

Pulse Shaping Filter Bank Example24 samples

in pulse

x marks samples of polyphase

filter

4 samples per symbol

Polyphase filter 3 response is the fourth sample of the

pulse shape plus every fourth sample after that

gTsym,3[n]

14 - 13

Symbol Clock Recovery

• Transmitter and receiver normally have different crystal oscillators

• Critical for receiver to sample at correct time instances to have max signal power and min ISI

• Receiver should try to synchronize with transmitter clock (symbol frequency and phase)First extract clock information from received signal

Then either adjust analog-to-digital converter or interpolate

• Next slides develop adjustment to A/D converter• Also, see Handout M in the reader

Optional

14 - 14

Symbol Clock Recovery

• g1(t) is impulse response of LTI composite channel of pulse shaper, noise-free channel, receive filter

)( )()()( 11*

symk

k kTtgatgtstq

k m

symsymmk mTtgkTtgaatqtp )( )( )()( 112

)(

)( )( } {)}({

21

2

11

symk

symk

symm

mk

kTtga

mTtgkTtgaaEtpE

Optional

ReceiveB()

SquarerBPFH()

PLLx(t)

q(t) q2(t)

p(t)

z(t)

E{ak am} = a2 [k-m]

g1(t) is deterministic

s*(t) is transmitted signal

Periodic with period Tsym

14 - 15

Symbol Clock Recovery

• Fourier series representation of E{ p(t) }

• In terms of g1(t) and using Parseval’s relation

• Fourier series representation of E{ z(t) }

tkj

kk

symeptpE )}({

sym symT tkj

symk dtetpE

Tp

0

)}({1

where

Optional

dkGGT

adtetg

T

ap sym

sym

tjk

symk

sym

11

221

2

2

ReceiveB()

SquarerBPFH()

PLLx(t)

q(t) q2(t)

p(t)

z(t)

dkGGT

akHkHpz sym

symsymsymkk 11

2

2

14 - 16

Symbol Clock Recovery

• With G1() = X() B()

Choose B() to pass ½sym pk = 0 except k = -1, 0, 1

Choose H() to pass sym Zk = 0 except k = -1, 1

• B() is lowpass filter with passband = ½sym

• H() is bandpass filter with center frequency sym

Optional

)cos(2 teeeZtzE symtjtj

k

tjkk

symsymsym

dkGGT

akHkHpZ sym

symsymsymkk 11

2

2

ReceiveB()

SquarerBPFH()

PLLx(t)

q(t) q2(t)

p(t)

z(t)